The necessity of the axiom of induction

Question

$\underline{First\ question}$
Let $P(n)$ be a proposition about $n$. In standard mathematical induction, we require:
(1)$P(0)$ holds.
(2)If $P(n)$ holds, $P(n+1)$holds.
Here we use "the axiom of induction" to prove $P(n)$ holds for all $n$. However, I can't figure out why "the axiom of induction" is necessary. For arbitrary $n\in \mathbf{N}$, $P(n)$ holds without the axiom in my opinion.

[edited from here]
If the above assumption (1) and (2) holds, for any given $n\in\mathbf{N}$, $P(2) = P(1+1), P(3)=P(2+1), \dots, P(n)=P((n-1)+1)$ holds.
[edited to here]

"Arbitrary" and "all" may be different, but if we write these two with logic symble,$\forall n\in\mathbf{N}$ ($P(n)$), the same. If we want to use the result of the proposition, say, $P(100)$, $P(100)$ evidently holds without the axiom, in my opinion. Why we have to prove $P(n)$ for "not any of $\mathbf{N}$" but "all of $\mathbf{N}$"?

$\underline{Second\ question}$
Please assume we have proved: $\forall n \in \mathbf{N}$, the number of elements in $\{1, 2, \ldots, n\}$ is finite. By induction, we have proved for all $n \in \mathbf{N}$ this proposition holds. Then, why $\mathbf{N}$ is an countable infinite set. I have know idea what is laid between finite and countable infinite.

Answer 1

First Question: You say "in your opinion". Try to prove it, then, and we will show the flaw.

Second Question: You proved every $I_n:=\{1,...,n \}$ is finite. "By induction", you just repeat it: you proved that for every $n \in \mathbb{N}$, the property holds, meaning: every $I_n:=\{1,...,n \}$ is finite. $\mathbb{N}$ is none of those $I_n:=\{1,...,n \}$, so why do you assume $\mathbb{N}$ is finite?

Answer 2

If you left out the axiom of induction from Peano's Axioms you could not eliminate the possibility that $x+1=x$ for some $x\in\mathbb{N}$, and other oddities that each of the other axioms would allow. There would also be no guarantee that if you started at $0$, you would eventually be able to reach any other element of $\mathbb{N}$ by repeatedly adding $1$. This may be more apparent with an alternative definition of induction:

$$\forall P\subset \mathbb{N}:[0\in P \land \forall x\in P: [x+1\in P]\implies P=\mathbb{N}]$$

Usually expressed in terms of a successor function $S$ on $\mathbb{N}$:

$$\forall P\subset \mathbb{N}:[0\in P \land \forall x\in P: [S(x)\in P]\implies P=\mathbb{N}]$$

Answer 3

First question

The point of induction is to convert the fact that you are able to prove $P(n)$ for any natural number $n$ into a proof that $P(n)$ for any natural number $n$. The two are 'very' different.

The first is:

For any natural number ( There is a proof that ( ... ) ).

The second is:

There is a proof that ( For any natural number ( ... ) ).

This is a switch in quantifiers which is invalid in almost all cases. Here also we cannot prove induction correct but can only argue that it is reasonable on certain grounds to accept it as true.

Second question $\def\nn{\mathbb{N}}$

Again you probably messed up the quantifiers. All you have proven is:

For any $n \in \nn$ ( $\{ k : k \in \nn \land k \le n \}$ is finite ).

Not:

$\{ k : k \in \nn \}$ is finite.